Higher Prelim Topic Revision - Web Version

Higher Formula Sheet

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1 Straight Line

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Hint

Perpendicular bisectors pass through the midpoint of a line segment, and run perpendicular to them.

Answer

\(4y=-3x+29\) or \(y=-\frac{3}{4}x+\frac{29}{2}\)

For more practice, try: Zeta Higher Textbook, Page 15, Exercise 1.8, Questions 1(a), 1(b) and 1(c)

Solution

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Hint

A straight line equation must be rearranged to make \(y\) the subject (\(y=\dots\)) before its gradient can be determined.

Answer

\(2y=-x-5\) or \(y=-\frac{1}{2}x-\frac{5}{2}\)

For more practice, try: Zeta Higher Textbook, Page 13, Exercise 1.6, Questions 3, 4 and 5

Solution

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Hint

The formula \(m=\tan{\theta}\) links a line's gradient to the angle it makes with the positive direction of the \(x\)-axis.

Answer

\(y=-x+2\)

For more practice, try: Zeta Higher Textbook, Page 9, Exercise 1.4A, Questions 1, 2 and 3

Solution

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Hint

Calculate the gradients of PQ and QR first. The conclusion to a collinearity problem needs to be learned carefully.

Answer

\(m_{PQ}=m_{QR}=-3\), valid statement

For more practice, try: Zeta Higher Textbook, Page 12, Exercise 1.5, Questions 1(a), 1(b) and 1(c)

Solution

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Hint

1. Medians meet opposite sides at the midpoint.

2. Altitudes meet opposite sides at an angle of 90 degrees.

3. You can use \(y=y\) as a first step to finding points of intersection.

Answer

1. \(y=-3x+5\)

2. \(2y=-x-5\) or \(y=-\frac{1}{2}x-\frac{5}{2}\)

3. \((3,-4)\)

For more practice, try: Zeta Higher Textbook, Page 21, Exercise 1.11B, Questions 1, 2 and 3

Solution (a)

Solution (b)

Solution (c)

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2 Recurrence Relations

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Hint

1. Calculate \(u_1\) first using the value of \(u_0\) as a starting point.

3. What is the value of \(a\)?

2. What is the formula for the limit of a recurrence relation?

Answer

1. \(15\)

3. \(-1<a<1\), valid statement

2. \(9\)

For more practice, try: Zeta Higher Textbook, Page 90, Exercise 6.1, Questions 1(a), 1(b) and 1(c)

Solution (a)

Solution (b)

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Hint

1. Can you put both \(u_2\) and \(u_3\) into the recurrence relation? Where does each one go?

2. What is the value of \(a\)?

Answer

1. \(-4\)

2. \(a>1\), valid statement

For more practice, try: Zeta Higher Textbook, Page 93, Exercise 6.3, Questions 1(f), 1(i) and 1(l)

Solution (a)

Solution (b)

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Hint

1. Substitute the value of \(u_4\) into the recurrence relation.

2. Use your expression from part (a), and the new information that \(u_5=-1\).

Answer

1. \(6k-4\)

2. \(k=\frac{1}{2}\)

For more practice, try: Zeta Higher Textbook, Page 15, Exercise 1.8, Questions 1(a), 1(b) and 1(c)

Solution (a)

Solution (b)

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Hint

1. What "12% is subtracted (from the original 100%) as a decimal?

2. Since dividing by \(0.12\) is tricky, multiply both parts of the fraction by \(100\) and work from there.

Answer

1. \(a=0.88,b-30\)

2. 250 squirrels

For more practice, try: Zeta Higher Textbook, Page 96, Exercise 6.5B, Questions 1, 2 and 3

Solution (a)

Solution (b)

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Hint

Create two equations: one linking the first two terms, and a second linking the second term and the third term. Solve simultaneously.

Answer

\(p=-\frac{1}{2},q=2\)

For more practice, try: Zeta Higher Textbook, Page 91, Exercise 6.2, Questions 1(a), 1(b) and 1(c)

Solution

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3 Differentiation I

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Hint

To differentiate, multiply by the power and then reduce the power by 1...

Answer

\(-19\)

Solution

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Hint

Start by splitting the fraction into two fractions, each with a denominator of \(3x^2\), and prepare each part for differentiation.

Answer

\(2x^2+2x^{-3}\)

Solution

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Hint

The gradient of a tangent to a curve is described by its derivative, \(\dfrac{dy}{dx}\).

Answer

\(9\)

Solution

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Hint

The gradient of a tangent was found in the last question. To find its equation, we also need to know a coordinate the tangent passes through.

Answer

\(y=6x-14\)

Solution

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Hint

The rate of change of a function is described by its derivative, \(h'(t)\). A square root \(\sqrt{x}\) can also be written in index as \(x^{\frac{1}{2}}\).

Answer

\(2\)

Solution

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Hint

Find the derivative, \(\dfrac{dy}{dx}\), first.

If a value of \(x\) is substituted into the derivative, the result will be the gradient. Here, when \(x=2\) is substituted in, what value would the output be equal to?

Answer

\(k=-\frac{1}{2}\)

Solution

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4 Quadratic Theory

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Hint

A few methods may be used.

One begins by fully expanding \(p(x+q)^2+r\).

Answer

\(3(x-4)^2+9\)

Solution

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Hint

A few methods may be used.

One begins by fully expanding \(a(x+b)^2+c\).

Answer

\(-(x-2)^2+7\)

Solution

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Hint

Start by considering the roots of the equation \(2x^2+8x-10\). Notice the common factor.

You need to draw a sketch.

Answer

\(x<-5,x>1\)

Solution

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Hint

Start by considering the roots of the equation \(m^2-m-20\).

You need to draw a sketch.

Since \(m\) is the variable, your answer must be in terms of \(m\).

Answer

Coming soon.

Solution

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Hint

An equation has (real) equal roots when \(b^2-4ac=0\).

Answer

\(p=\pm6\)

Solution

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Hint

An equation has no real roots when \(b^2-4ac<0\).

Answer

\(q<-\frac{6}{5}\)

Solution

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Hint

An equation has real, distinct roots when \(b^2-4ac>0\).

Answer

\(k<-2,k>6\)

Solution

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Hint

Points of intersection can be found by solving \(y=y\)...

Answer

\((-2,1)\) and \((4,5)\)

Solution

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Hint

Tangency can be explored by solving \(y=y\) are for generally finding points of intersection, then considering how the resulting equation can be factorised or \(b^2-4ac\).

Answer

\((2,5)\)

Solution

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5 Sets and Functions

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Hint

Start by writing it in the form \(y=...\) and then rearrange to make \(x\) the subject.

Answer

\(h^{-1}(x)=3x+4\)

Solution

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Hint

Start by writing it in the form \(y=...\) and then rearrange to make \(x\) the subject.

Answer

\(f^{-1}(x)=\left(\dfrac{x+3}{2}\right)^2\)

Solution

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Hint

Division by zero is undefined.

Answer

\(x\ne \frac{1}{2}\)

Solution

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Hint

The square root of a negative is undefined.

Answer

\(x<3\)

Solution

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Hint

The expression for \(g(x)\) needs to be substituted into the expression for \(f(x)\).

Answer

\(f(g(x))=2x-1\)

Solution

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Hint

The expression for \(f(x)\) needs to be substituted into the expression for \(g(x)\).

Simplify it fully.

2. Applying \(f\) to \(x\) and then applying \(g\) to the result leads back to \(x\). What is happening here?

Answer

1. \(g(f(x))=x\)

2. \(f\) and \(g\) are inverse to each other.

Solution

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Hint

1. Substitute \(g(x)\) into \(f(x)\) and simplify.

2. Observe the results from part (a) before simplifying, when it when written in completed square form, \((x-a)^2+b\). What is the lowest possible value of \(k(x)\)?

3. Begin by factorising \(k(x)\).

Answer

1. \(k(x)=x^2-6x+8\)

2. \(k(x)\geqslant-1\).

Solution

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6 Trigonometry

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Hint

Rearrange to make \(\sin x\) the subject. Give your final answer in radians.

Answer

\(x=\frac{\pi}{6},\frac{5\pi}{6}\)

Solution

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Hint

Use a CAST diagram carefully to identify quadrants in which \(\tan x\) is negative.

Answer

\(x=\frac{5\pi}{12},\frac{11\pi}{12}\)

Solution

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Hint

1. At which angle is \(y=4\cos x\) at its minimum, and how low does it go? Answer in radians.

2. The points of intersection Q and R can be obtained by solving the set of equations \(y=4\cos x\) and \(y=2\) simultaneously.

Answer

1. P\((\pi,-4)\)

2. Q\(\left(\frac{\pi}{3},2\right)\), R\(\left(\frac{5\pi}{3},2\right)\)

Solution

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Hint

For a \(2x\) trig equation from 0 to 360 degrees, expect 4 solutions.

Answer

\(x=30,150,210,330\)

Solution

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Hint

For a \(2x\) trig equation from 0 to 360 degrees, expect 4 solutions.

Answer

\(x=120,150,300,330\)

Solution

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Hint

Remember that \(\frac{\pi}{6}\) radians is equal to \(30\degree\).

Find \(x\degree-30\degree\) solutions then add \(30\degree\) to solve for \(x\).

Answer

\(x=\frac{11\pi}{12},\frac{23\pi}{12}\)

Solution

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Hint

Expect 4 solutions... Use a combination of skills from the previous questions.

Answer

\(x=\frac{\pi}{4},\frac{5\pi}{12}.\frac{5\pi}{4}.\frac{17\pi}{12}\)

Solution

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7 Graph Transformations

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Hint

Make a list of all coordinates on the original graph (here there are 3).

Everything outside the \((x)\) bracket affects \(y\)-coordinates. These transformations are usually seen as intuitive.

Everything inside the \((x)\) bracket affects \(x\)-coordinates. These transformations are usually seen as counterintuitive.

Make sure your final graph has all the transformed images of the original coordinates clearly shown.

Answer

See full solution below for the graph and coordinates.

Solution

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Hint

Remember to apply the two transformations following the rules of operation: \(\times(-1)\) then \(-1\)

As well as changing the coordinates, think about how the transformations will change the shape and position of the graph. Here, \(-f(x)\) will flip the graph upside down.

Answer

See full solution below for the graph and coordinates.

Solution

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Hint

If \(-f(x)\) is a vertical flip, what is \(f(-x)\)?

Answer

See full solution below for the graph and coordinates.

Solution

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Hint

Don't ignore the coordinate of \((0,0)\) on the original graph.

Answer

See full solution below for the graph and coordinates.

Solution

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Hint

\(2-g(x)\) can be rewritten as \(-g(x)+2\).

Answer

See full solution below for the graph and coordinates.

Solution

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Hint

\(\frac{1}{2}g(x)\) is a vertical compression by factor 2.

Answer

See full solution below for the graph and coordinates.

Solution

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8 Vectors

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Hint

Find the components of vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{QR}\).

Make sure your collinearity statement is carefully and fully worded.

Draw a sketch to make (b) easier to see.

Answer

1. \(2\overrightarrow{PQ}=\overrightarrow{QR}\), valid statement.

2. \(1:2\)

Solution (a)

Solution (b)

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Hint

Check which method your teacher taught you to use. One option is the section formula.

Answer

E\((5,-1,4)\)

Solution

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Hint

1. Use \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\)

2. The formula sheet reminds you how to do this.

3. Use the formula \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos{\theta}\), rearranged and in terms of \(\overrightarrow{EM}\) and \(\overrightarrow{ER}\).

Answer

1. \(\overrightarrow{EM}\)=

   \[\begin{pmatrix} \phantom{-}2\\ -6\\ -1 \end{pmatrix}\]

   and \(\overrightarrow{ER}=\)

   \[\begin{pmatrix} -3\\ -2\\ \phantom{-}3 \end{pmatrix}\]

2. 3

3. \(84.3^{\circ}\)

Solution (a)

Solution (b)

Solution (c)

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Hint

If a vector has a magnitude of \(k\), multiplying that vector by \(\frac{1}{k}\) will create a parallel vector with a length of \(1\) (a unit vector).

Answer

\(\begin{pmatrix} \frac{1}{3}\\ -\frac{2}{3}\\ \frac{2}{3} \end{pmatrix}\)

Solution

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Hint

1. Start by expressing vector BF as BA\(+\)AC\(+\)CF.

2. Start similar to the above hint.

3. The following approaches will help:

• For all vectors, \(\mathbf{a}\cdot\mathbf{a}=|\mathbf{a}|^2\)

• If \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular then \(\mathbf{a}\cdot\mathbf{b}=0\)

• If the angle \(\theta\) between two vectors and their magnitudes are known, then \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos{\theta}\) can be used.

Answer

1. \(-\mathbf{u}+\mathbf{v}+\mathbf{w}\)

2. \(-\frac{2}{3}\mathbf{w}-\mathbf{v}+\mathbf{w}\)

3. 1. \(\frac{9}{2}\) (ii) \(\frac{9}{2}\)

Solution (a)

Solution (b)

Solution (c)

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9 Differentiation II

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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10 Polynomials

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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11 Integration

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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12 Addition Formulae

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Hint

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Answer

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Solution (a)

Solution (b)(i)

Solution (b)(ii)

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Hint

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Answer

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Solution

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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13 The Circle

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution

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Hint

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Answer

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Solution